(i) The vector \(\overrightarrow{XP} = -4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}\). For \(\angle OPX = 90^0\), the dot product \(\overrightarrow{OP} \cdot \overrightarrow{XP} = 0\). This gives:

\([-4\mathbf{i} + (p-5)\mathbf{j} + 2\mathbf{k}] \cdot (pj + 2k) = 0\)

\(p^2 - 5p + 4 = 0\)

Solving the quadratic equation, \(p = 1 \text{ or } 4\).

(ii) For \(p = 9\), \(\overrightarrow{XP} = -4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}\). The magnitude \(|\overrightarrow{XP}| = \sqrt{16 + 16 + 4} = 6\).

The unit vector is \(\frac{1}{6}(-4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k})\).

(iii) \(\overrightarrow{AG} = -4\mathbf{i} + 15\mathbf{j} + 2\mathbf{k}\). Since \(\overrightarrow{XQ} = \lambda \overrightarrow{AG}\), and \(\lambda = \frac{2}{3}\),

\(\overrightarrow{XQ} = -\frac{8}{3}\mathbf{i} + \frac{10}{3}\mathbf{j} + \frac{4}{3}\mathbf{k}\).